Circuit Analysis Techniques
We learned that KVLs, KCLs, and i-v characteristics equations results in a set of linear
equations for the circuit variables (typically 2E equations in 2E circuit variables where E is
number of elements). We learned that we can use i-v characteristics equations when we are
marking the circuit variables and reduce the number of equation to be solved to E equations
in E unknowns (which is the same as number of KCLs plus KVLs). In this section, we
learn two methods, node-voltage and mesh-current, which reduce the number of equations
to either number of KCLs or number of KVLs. These are node-voltage and mesh-current
methods. They also form the basis to deduce certain properties of linear circuits.
Node-Voltage Method
The node-voltage method is based on following idea. Instead of solving for circuit variables,
i and v of each element, we solve for a different set of parameters, node voltages in this case,
which automatically satisfy KVLs. As such, we do not need to write KVLs and only need
to solve KCLs.
+
Recall definition of potential. Potential was defined as
the voltage between any point and the reference point
(usually called ground or common). Take a circuit and
identify the nodes on the circuit. Define the node voltage as the potential of that node (voltage between that
node and the reference). In the circuit shown, the node
voltages are denoted as v1 , v2 , and v3 . The voltage across
each element is simply the voltage of the node attached
to + terminal of the element minus the voltage of the
node attached to − terminal of the element (see page 1
of notes):
v1
va
+
v2
+
vb
vc
−
−
Reference Node
(Ground)
_
v3
va = v1 − v2
vb = v1 − v3
vc = v2 − v3
Note that using node voltage definitions, KVLs will be automatically satisfied:
KVL :
−vb + va + vc = 0
−(v1 − v3 ) + (v1 − v2 ) + (v2 − v3 ) = 0
MAE140 Notes, Winter 2001
→
0=0
25
As the choice of the reference point (ground) is arbitrary, we can choose the reference point
to be any node on the circuit. For example, we could choose node 3 in the above circuit as
the reference mode (v3 = 0). The, number of node voltages are reduced to 2. the voltage
across the three elements are
va = v1 − v2
vb = v1
vc = v2
and KVL is still automatically satisfied. With this simplification, we have N − 1 node
voltages in each circuit where N is the number of nodes.
Recall that number of independent KCLs is N − 1. If we can write the current in each
element in terms of the node voltages, then the circuit reduced to N − 1 KCLs in N − 1 node
voltages. This is straightforward for two of the three linear elements we have encountered
up to now:
Current Source: The current flowing through current source is independent
of voltage across it and, therefore, is independent of node voltages. From the
figure, it is obvious that the current leaving node 1 is i1 = is and current
leaving node 2 is i2 = −is .
v1
is
−
v2
Voltage Source: The voltage across a voltage source is constant and is independent of current flowing through it. The i-v characteristics of the voltage
source cannot be used to find the current flowing through it. Voltage sources
should be treated in a special manner that will be discussed later.
Resistor: Resistors follow Ohm’s Law. We need to calculate the voltage
across the resistor first and find the current using the Ohm’s Law. Care must
be taken as we should use passive sign convention. For example, if we want to
find current leaving node 1, i1 , we first find voltage v12 :
v12 = v1 − v2
→
i1 =
i1
+
vs
i2
v1
−
+
v21 R
+
−
i2
v12
v1 − v2
=
R
R
i1
v12
v2
Alternatively, if we want to find the current leaving node 2, i2 , we calculate v21 first:
v21 = v2 − v1
→
i2 =
v2 − v1
v21
=
R
R
Thus, if a resistor is connected between nodes 1 and 2, the current flowing from node 1 to 2
v2 − v1
v1 − v1
is
and the current flowing from node 2 to 1 is
R
R
MAE140 Notes, Winter 2001
26
In summary, we have found that we can define node voltages as circuit variables and:
1. Voltage across each element can be written in terms of node voltages.
2. Current through each element (resistor and ICS) can be written in terms of node voltages
using the i-v characteristics of that element.
3. KVLs are automatically satisfied.
What is left is a set of KCLs. By using node voltages, we have reduced the number of
variables to N − 1 node voltages (one node is the reference node with zero voltage) and
N − 1 KCLs to solve.
Example:
The circuit has three nodes. We choose the node
at the bottom as the reference (ground). The voltages at the other nodes are denoted by v1 and v2 .
We write KCL at these two nodes:
KCL 1:
i2 + i1 − is1 = 0
→
KCL 2:
is2 + i3 + i2 = 0
→
i2
v
1
R2
i’2
i3
i1
R1
v2
is2
R3
is1
v1 − v2 v1 − 0
+
= is1
R2
R1
v2 − 0 v2 − v1
+
= −is2
R3
R2
The above are two equations in two unknowns (node voltages v1 and v2 ). Note that a circuit
with 5 elements which, in principle, has 10 circuit variables and 10 equations to be solved is
reduced to 2 equations in 2 unknowns.
The previous equations can be rearranged as:
1
1
1
+
− v2
= is1
v1
R1 R2
R2
1
1
1
−v1
+ v2
+
= −is2
R2
R2 R3
or using the definition of conductance, G = 1/R, we have:
(G1 + G2 )v1 − G2 v2 = is1
−G2 v1 + (G2 + G3 )v2 = −is2
In the matrix form, the equations are:
G1 + G2
−G2
−G2
G2 + G3
•
v1
v2
MAE140 Notes, Winter 2001
=
is1
−is2
27
This can be generalized to any circuit with N nodes (let n = N − 1). The node-voltage
equation can be written as
G • v = is

G=





G11 −G12
−G21
G22
...
...
−Gn1 −Gn2
... −G1n
... −Gnn
...
...
...
Gnn












v=
v1
v2
...
vn












is =
is1
is2
...
isn






Matrix G is called the conductance matrix. If the KCLs are written as sum of current
exiting each node, the diagonal elements of this matrix will be positive and the off-diagonal
elements will be negative. The diagonal elements Gii are equal to the sum of all conductances
connected to node i. The matrix is symmetric, i.e. Gij = Gji. These off-diagonal elements,
Gij , are sum of conductances directly connecting nodes i and j. Array v is the array of node
voltages. Array is is the array of current sources. Element isi is net equivalent of source
currents entering the node.
Because of the above features of the node-voltage equations, it is possible to directly write
the node voltage equations in the matrix form of G • v = is and follow the above rules to
construct matrix G and vectors v and is . Node voltage method is a powerful and simple
method. As such, all circuit simulation computer code such as PSpice use node-voltage
method and use the above procedure to construct the conductance matrix and the nodevoltage equations.
I personally have found that in solving circuits by hand, it is much better to forgo the above
direct technique for writing matrix G and vector is and instead just write down the KCLs
as sum of currents existing a node. The chances of mistakes is much reduced and is worth
the price of one additional step of rearranging terms to arrive at the above matrix equation.
Furthermore, the above procedure should be modified for circuits containing voltage sources.
It is just not worth memorizing two additional procedures.
MAE140 Notes, Winter 2001
28
Example: In the circuit below, resistance are given in Ohms. Find i
5A
v
v2
1
i
v3
6
12
7A
4
17 A
4
12
3
The circuit has four nodes. Assigning the node at the bottom as the reference node, we write
KCL at the three other nodes:
Node 1:
Node 2:
Node 3:
v1 − v2 v1 − 0
+
−7=0
12
4
→
4v1 − v2 = 24
(v1 − v2 ) + 3v1 − 24 = 0
v2 − v3 v2 − 0 v2 − v1
+
+
−5 =0
6
4
12
→
−v1 + 6v2 − 2v3 = 60
2(v2 − v3 ) + 3v2 + (v2 − v1 ) − 60 = 0
v3 − 0 v3 − 0 v3 − v2
−17 +
+
+
=0
3
12
6
−204 + 4v3 + v3 + 2(v3 − v2 ) = 0
→
−2v2 + 7v3 = 204
+5 +
or




4v1 −v2
= 24
−v1 +6v2 −2v3 = 60



−2v2 +7v3 = 204
Note the structure of above matrix equations. The G matrix is symmetric. As we wrote the
KCLs as the sum of currents exiting the nodes, the diagonal elements are positive and offdiagonal elements are negative. These are good points to check to see if we made a mistake
in writing KCLs.
The above equations can be solved to get v1 = 12 V, v2 = 24 V, and v3 = 36 V. We know
need to find the problem unknown, i from the node voltages. Using Ohm’s law, we get
i=
v2 − v3
24 − 36
=
= −2 A
6
6
MAE140 Notes, Winter 2001
29
Notes:
1. The above circuit has 8 circuit elements. Node-voltage method results in only 3 equations
in 3 unknown. You can try to solve the above circuit with KVL/KCL or circuit reduction
techniques to appreciate the power and utility of the node-voltage method.
2. In the above circuit, the 12 and 3 Ω resistors are in parallel. The number of node-voltage
equations would not have changed if we had replaced these two parallel resistors with their
equivalent. So, as a matter of principle, there is no need to replace parallel elements if
you are using node-voltage method. It pays, however, to reduce series element as this will
eliminate the node between the series elements and reduces the number of equations.
Circuits with voltage sources
Node voltage method as outlined above has to be modified for circuits with voltage sources
as the i-v characteristics of IVS does not specify the current through it. This modifications
is best explained in the context of the example below:
Example: Find v.
v1
+
−
6 kΩ
3V
v2
+
The circuit has four nodes. We will assign
the node at the bottom as the reference node
and we have three node voltages as unknowns.
If we did not have voltage sources, we would
write 3 KCL at the 3 nodes to arrive at 3 equations in 3 unknowns. For this circuit, writing
KCLs is not useful as the i-v characteristics
of IVS does not specify the current through
it. We note that there are two possible configuration for voltage sources:
20 V
2 kΩ
v3
−
+
v
4 kΩ
6 mA
−
a) Voltage source is connected to the reference node (such as the 20-V IVS in the circuit
above). If we write the voltage across the IVS in terms of the node voltages:
v1 − 0 = 20
→
v1 = 20 V
In this case, we do not need to write any KCL. The value of node voltage can be derived
directly from the i-v characteristics of the IVS!
b) Voltage source is connected between two nodes that are not reference nodes (such as the
3-V IVS in the above circuit). We need to write 2 equations for nodes 2 and 3 (one for each).
KCLs for those nodes (2 equations) cannot be used as both contain the current through IVS
MAE140 Notes, Winter 2001
30
that is unknown. One equation is the voltage across the IVS in terms of the node voltages:
v2 − v3 = 3
The second equation can be found by noting that as charge cannot be accumulated in any
part in the circuit, KCL should apply to any portion of the circuit. Consider a portion of the
circuit including nodes 2, 3 and the 3 V voltage source. We call this portion a supernode.
KCL should be valid for a supernode. This gives
KCL for super node 2&3:
v3
v2
v2 − v1
+
+
=0
3
3
4 × 10
2 × 10
63
−72 + 3v3 + 6v2 + 2(v2 − v1 ) = 0
−6 × 10−3 +
−2v1 + 8v2 + 3v3 = 72
We now have three equations in three unknowns:







v1
−2v1
= 20
v2 −v3 = 3
+8v2 +3v3 72
Which gives v1 = 20 V, v2 = 11 V, and v3 = 8 V. The problem unknown v = v3 − 0 = v3 =
8 V.
KCL for super node 2&3:
−6 × 10−3 +
Supernode
v = 20 V 6 kΩ
1
3V
v2
+
Note that most of the work outlined above
could be done on the circuit diagram as
shown. Noting v1 = 20, we can write the
value of the node voltage on the circuit. Also,
using v2 − v3 = 3, we can arrive at v3 = v2 − 3
as shown in the circuit below.
Now, we only have one unknown, v2 , which
can be found by writing KCL on the super
node:
+
−
20 V
i
v3 = v2 − 3
+
v
4 kΩ
−
2 kΩ
6 mA
−
v2 − 3
v2
v2 − 20
+
+
=0
3
3
4 × 10
2 × 10
63
The above is one equations in one unknown which can be solved rapidly.
MAE140 Notes, Winter 2001
31
Notes:
1. Voltage sources actually simplify the node voltage equations. By using the i-v characteristic equations of IVS on the circuit diagram, we can reduce the number of unknown
node voltages (an, thus, number of equations to be solved) to N − 1 − NIV S .
2. The situation is much simpler for voltage sources that are connected between a node
and ground (such as 20-V source in the above circuit). As the location of the reference
node is arbitrary, a good rule is: Choose the reference node as the node with maximum
number of voltage sources attached to it!
3. While we do not need to write KCL at the nodes connected to a voltage node in order
to find the node voltages, we need to use KCL if we want to find the current through
a voltage source. For example, in the circuit above, current i can be found by writing
KCL at node 3 (after we have found all node voltages):
v2 − 3
=0
4 × 103
11 − 3
= −6 + 2 = −4 mA
i = −6 × 10−3 +
4 × 103
−i − 6 × 10−3 +
4. In writing Node voltage equations (KCLs), one should ignore the reference directions
provided on the circuit for circuit variables (e.g. i and v in the above circuit). The node
voltage equations are written in terms of node voltages following the above procedure
and are independent of these reference directions.
Recipe for Node Voltage Method
1. Identify nodes and supernodes.
a) Choose the reference node as the one with maximum number of voltage sources
attached to it.
b) Use i-v characteristic equations of IVS to find node voltage values and reduce the
number of unknowns
2. Write KCL at each node or supernode.
3. Solve node-voltage equations.
4. Calculate problem unknowns from node voltages. If you need to calculate the current
in a voltage source you may have to write KCL at a node connected to that voltage
source.
MAE140 Notes, Winter 2001
32